A STUDY OF HYDRAULIC FRACTURE INITIATION IN TRANSVERSELY ISOTROPIC ROCKS. A Thesis VAHID SERAJIAN

A STUDY OF HYDRAULIC FRACTURE INITIATION IN TRANSVERSELY ISOTROPIC ROCKS A Thesis by VAHID SERAJIAN Submitted to the Office of Graduate Studies of Texas A&M University in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE August 2011 Major Subject: Petroleum Engineering

A Study of Hydraulic Fracture Initiation in Transversely Isotropic Rocks Copyright 2011 Vahid Serajian

A STUDY OF HYDRAULIC FRACTURE INITIATION IN TRANSVERSELY ISOTROPIC ROCKS A Thesis by VAHID SERAJIAN Submitted to the Office of Graduate Studies of Texas A&M University in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE Approved by: Chair of Committee, Committee Members, Head of Department, Ahmad Ghassemi Stephen A. Holditch Benchun Duan Stephen A. Holditch August 2011 Major Subject: Petroleum Engineering

iii ABSTRACT A Study of Hydraulic Fracturing Initiation in Transversely Isotropic Rocks. (August 2011) Vahid Serajian, B.S.; M.S., Amirkabir University of Technology (Tehran Polytechnic) Chair of Advisory Committee: Dr. Ahmad Ghassemi Hydraulic fracturing of transverse isotropic reservoirs is of major interest for reservoir stimulation and in-situ stress estimation. Rock fabric anisotropy not only causes in-situ stress anisotropy, but also affects fracture initiation from the wellbore. In this study a semi-analytical method is used to investigate these effects with particular reference to shale stimulation. Using simplifying assumptions, equations are derived for stress distribution around the wellbore s walls. The model is then used to study the fracture initiation pressure variations with anisotropy. A sensitivity analysis is carried out on the impact of Young s modulus and Poisson s ration, on the fracture initiation pressure. The results are useful in designing hydraulic fractures and also can be used to develop information about in-situ rock properties using failure pressure values observed in the field. Finally, mechanical and permeability anisotropy are measured using Pulse Permeameter and triaxial tests on Pierre shale.

iv DEDICATION To my parents, Mohammadreza Serajian and Mahin Zinsazi and my siblings, Sanaz and Sahand Serajian, for all their support and help

v NOMENCLATURE a Radius of the well a ij Components of compliance tensor E, E Young s modulus in parallel and perpendicular planes E h, E v h i, h j Young s modulus in horizontal and vertical planes, respectively Direction cosines of the i and j axes in relation to the axis of symmetry H Compliance matrix in anisotropy coordination system I K K h, K v l,m,n Imaginary unit Anisotropy ratio, permeability Permeability in horizontal and vertical directions, respectively Direction cosines of the unit vectors in x, y, z directions respectively P w Mud pressure in the well P p Pore pressure e Real component of a complex number Biot effective stress coefficient, h v Biot effective coefficient in horizontal and vertical directions ij Reduced strain coefficient

vi h, H Strain component in min. and max. principal horizontal stress directions ij Kronecker delta Poro elastic coefficient ( 0 1) i 1,2,3 Characteristic roots of the stress i ', i 1,2,3 Derivatives of the stress analytical functions i v Vertical stress h Minimum horizontal stress H Maximum horizontal stress t Rock tensile strength eff Effective stress xx, yy, zz Total normal stresses in Cartesian coordination xo,, yo,, zo, Original (in-situ) stresses in Cartesian coordination xi,, yi,, zi, Induced stresses in Cartesian coordination xy, xz, xy Total shear stresses in Cartesian coordination xy, o, xz, o, xy, o Original (in-situ) shear stresses in Cartesian coordination, ' Poisson s ration in the parallel and perpendicular planes, respectively

vii h, v Poisson s ration in horizontal and vertical planes, respectively

viii TABLE OF CONTENTS Page ABSTRACT... DEDICATION... NOMENCLATURE... TABLE OF CONTENTS... LIST OF FIGURES... LIST OF TABLES... iii iv v viii x xii 1. INTRODUCTION AND LITERATURE REVIEW... 1 2. GENERAL DEFINITIONS, STRESS AND STRAIN EQUATIONS... 3 2.1 Introduction... 3 2.2 Basic Definitions... 4 2.3 Sign Convention... 5 2.4 Constitutive Relations... 6 2.5 Concept of Transverse Isotropy... 10 2.6 Pore Pressure and Effective Stress... 12 2.7 Transformation of Elastic Constants... 13 3. ANALYTICAL EQUATIONS FOR FRACTURE INITIATION PRESSURE. 16 3.1 Introduction... 16 3.2 General Expressions for Calculation of the Stress Functions... 17 3.3 Stress State around the Wellbore... 21 3.3.1 Wellbore Stresses in Transversely Isotropic Rock in Generic Form... 22 3.3.2 Wellbore Stresses in Transversely Isotropic Rock Drilled Along Principal in-situ Stresses... 28 3.4 Assessment of a Proper Failure Criterion for Fracturing Analysis... 31 3.5 Estimation of Fracture Initiation Pressure... 33

ix 4. CALCULATION OF FRACTURE INITIATION PRESSURE IN A HORIZONTAL WELL DRILLED IN SHALE... 35 Page 4.1 Introduction... 35 4.2 Numerical Example to Find Tangential Stress around the Well... 35 4.3 Comparison around the Well Using Isotropic and Anisotropic Solution... 41 4.4 Comparison around the Well Using Different Anisotropy Ratios. 43 4.5 Estimation of Fracture Initiation Pressure in Horizontal Wells Drilled in Horizontally Deposited Rock... 45 4.6 Direct and Indirect Estimation of Rock Anisotropy using Experimental Tests... 56 4.7 Variations in Fracture Initiation Pressure Caused by Well Orientation and Rock Anisotropy... 61 5. CONCLUSIONS AND SUMMARY... 71 REFERENCES... 72 APPENDIX 1... 75 APPENDIX 2... 84 VITA... 94

x LIST OF FIGURES Page Fig. 1 Three dimensional stress notations.... 6 Fig. 2 Axes orientation in a transversely isotropic material.... 10 Fig. 3 Position of the horizontal well, local and global coordinate systems attached in a transverse isotropic rock.... 14 Fig. 4 Transverse isotropy in a plane perpendicular to the well axis... 20 Fig. 5 Transverse isotropy in a plane parallel to the well axis... 21 Fig. 6 Transformation around the wellbore from rectangular to polar system... 25 Fig. 7 Horizontal well orientation with respect to 3 coordination systems... 29 Fig. 8 Well position with respect to in-situ stresses and the coordination system.... 36 Fig. 9 distribution around a nearly isotropic horizontal well drilled along H max and with mud pressure of 8000 psi using isotropic and anisotropic solutions. 42 Fig. 10 Tangential stress distribution around the well for three cases with different anisotropy ratios.... 44 Fig. 11 Fracturing pressure variations in a well drilled along H max by changing Young s modulus and Poisson s rations.... 48 Fig. 12 Fracturing pressure variations in a well drilled along hmin by changing Young s modulus and Poisson s rations.... 49 Fig. 13 Variations in equivalent mud weight required to cause fracturing in rocks with different anisotropy ratios.... 53 Fig. 14 Fracturing pressure variation by changing shear modulus and Poisson s ration... 55

xi Page Fig. 15 Respective orientation of rock bedding and the core samples in Pierre shale rock.... 57 Fig. 16 Comparison of Young s modulus for samples cored in 2 orthogonal directions in Pierre shale rock.... 58 Fig. 17 Comparison of Poisson s ration for samples cored in 2 orthogonal directions in Pierre shale rock.... 59 Fig. 18 Stress dependent permeability for Pierre shale samples cored in parallel and perpendicular directions.... 60 Fig. 19 Three different coordination systems in the anisotropic rocks.... 63 Fig. 20 Orientation of local coordination system vs. the global coordination (principal stresses direction)... 64 Fig. 21 Orientation of the rectilinear anisotropy plane vs. the global coordination system (principal stresses direction)... 65 Fig. 22 Fracturing pressure variations caused by well orientation from H max anisotropy ratio Ehor K.... 67 E vert Fig. 23 Variations of horizontal well angle from Z to X axes directions.... 69 Fig. 24 Orientation of global coordination system vs. the coordination system attached to the plane of rectilinear anisotropy... 79 Fig. 25 Orientation of global vs. local coordinate systems... 82 Fig. 26 Schematic of pressure pulse permeameter... 88 and

xii LIST OF TABLES Page Table 1 Available in-situ Stresses and Mechanical Properties for Calculations... 36 Table 2 Mechanical Properties of the Rock Used to Validate the Solutions... 37 Table 3 Mechanical Properties of the Rock Being Used for Sensitivity Analysis... 46 Table 4 Numerical Example to Show Variation of Fracturing Pressure with Different Young s Modulus and Poisson s Ration... 47

1 1. INTRODUCTION AND LITERATURE REVIEW Unconventional petroleum resources are among the most important sources of energy and tend to occur in formations with elastic and hydraulic anisotropy. The different mechanical properties in different directions can cause difficulty in accurately estimating the safe mud weight while drilling and the pressure required for stimulation during fracturing treatment. Consideration of general anisotropy is impractical, but the commonly encountered case of transverse isotropy lends itself to analytical treatment. A transversely isotropy rock is one with mechanical properties that are symmetric about an axis (called axis of rotation). Laminated sedimentary rocks such as shales can be classified as transversely isotropic. Very low permeability is the main constraining factor in gas production from shales so that stimulation from inclined or horizontal wells becomes necessary. Prediction of fracture initiation pressure in inclined or horizontal wellbores is essential for safe drilling and efficient hydraulic fracture stimulation. To predict the pressure requirements in these activities, the stress distribution around the wellbore should be assessed. This is often achieved using (Kirsch 1898) solution for elastic and isotropic rocks. (Haimson and Fairhurst 1967) proposed analytical equations for stress state around a vertical borehole in elastic rock for estimating the fracturing pressure. This thesis follows the style of SPE Journal.

2 (Hossain et al. 2000) also estimated the fracture initiation pressure for open hole wells drilled in isotropic rocks with different trajectories. Other equations have also been proposed for stress distribution around a wellbore in anisotropic elastic and poro elastic rock. (Deily and Owens 1969; Hsiao 1988; Jaeger et al. 2007) Stress analysis assuming isotropy can be inaccurate and often underestimate fracturing pressure (e.g., Suarez-Rivera et al., 2006). (Amadei 1983) and (Lekhnitskii 1963) solved the stress distribution around inclined boreholes in transverse isotropic rocks. Aadnoy (Aadnoy 1988; Aadnoy 1989; Aadnoy and Chenevert 1987) simplified Amadei and Lekhnitskii s methods to estimate the stress distribution around horizontal wells but he neglected the effect of Poisson s ration difference in vertical and horizontal directions for transversely isotropic rocks. His simplified equations may result in erroneous conclusions, especially during back analysis to estimate the in-situ stresses and rock mechanical parameters. Ong et al. (Ong 1994; Ong and Roegiers 1993; Ong and Roegiers 1996) also implemented (Amadei 1983) and (Lekhnitskii 1963) method for wellbore stability and fracture initiation estimation in inclined and horizontal wells. (Abousleiman and Cui 1998) solved the stress distribution around a well in transversely isotropic poroelastic rock drilled perpendicular to the plane of isotropy. In this study, an analytical technique proposed by (Lekhnitskii 1963) and (Amadei 1983) is used to study the stress distribution around wellbore drilled horizontally in a transversely isotropic formation. The focus is on linear elasticity and the time dependent and plastic behaviors of rock are not considered.

3 2. GENERAL DEFINITIONS, STRESS AND STRAIN EQUATIONS 2.1 Introduction A rock mass is subjected to a combination of gravitational and tectonic stresses. Knowledge of the stress distributions is critical in geosciences and geo-engineering projects since it is the most important factor in assessing the stability of the underground openings. Different rocks with various mechanical properties exist in the nature. The most commonly used procedure in the engineering projects assumes the rock mass as a CHILE 1 rock. Assuming the CHILE rock in real projects does not seems to be the perfect choice in all the cases since anisotropy, heterogeneity and nonlinearity is always seen in all rocks and should be taken into account in real world. Shales are one of the most important examples of such cases that assuming their mechanical properties as isotropic may lead to wrong interpretations in the projects especially in deep reservoirs with high pressures and challenging conditions. In this section, the mechanical properties of the shales are reviewed and the effects of different foliation, in-situ stress orientation and the wellbore trajectory are studied on the overall compliance matrix of the stress-strain relation. The results will be used for studying the hydraulic fracturing initiation in horizontal wells in shales. 1 Continuous Homogeneous Isotropic Linearly Elastic

4 2.2 Basic Definitions There are some important definitions which should be mentioned before starting the discussions; - Elasticity: A rock is called Elastic when the deformations associated with loading are all recoverable after the unloading. - Homogeneity, Heterogeneity: The rock is called homogeneous if the mechanical properties are identical in different locations and is heterogeneous if these properties vary with the location from point to point. - Orthotropic Materials: A material that possesses different mechanical properties in different orthogonal directions. - Isotropy, Anisotropy: A rock is called isotropic when its mechanical properties don t change with the direction and is called anisotropic when these properties change with direction. Anisotropy is the characteristic of the rocks with long minerals and layered deposition. Shales, schists, slates and bedded sandstones are the most typical anisotropic rocks with layered structure. In case of sandstones, they are usually considered as isotropic but in some of them due to the mixture of clay minerals in the rock mass, existence of micro fracs in some orientations and some other structural characteristics, the behavior of the sandstones cannot be explained by isotropy and should be considered anisotropic.

5 2.3 Sign Convention (Jaeger et al. 2007) suggest that in stress and strain analyses in the rocks, compressive stresses should be taken as positive. They mention that the compressive stresses are more common in rock mechanics than the tensile stresses. If we are to use the same sign convention as most other engineering (i.e. mechanical engineering), we should have added a negative sign to all the equations that deal with compressive stresses which finally increases the complexity in the analyses. In this study: Normal stresses are shown with Sigma sign (i.e. xx ) Shear stresses are shown with tau sign (i.e. xy ) The stress notations of the compressive and shear stresses are shown in Fig. 1. The coordinate system and the stress directions are also shown accordingly. Since this study is sensitive to the stress direction, knowledge of the sign convention is essential.

6 Fig. 1 Three dimensional stress notations. As can be inferred from Fig. 1, the first subscript of the stresses shows the surface which the stress vector acts and the second subscript shows the direction of the stress component. For example, the shear stress xy indicates that the stress is in X plane and in Y direction. Other stress components can be explained the same way. 2.4 Constitutive Relations In engineering, a constitutive equation is a relationship between the stress and strain in a specific material. In other words, constitutive equation can approximate the occurring strain of a material which is under stress. The constitutive equations can be linear, if the deformations are relatively small and can be non-linear in large displacements such as collapses and deteriorations. In this study, since we are dealing

7 with comparatively small deformations, the linear constitutive equations may be the most reasonable assumption. To calculate the response of a linear elastic homogeneous and continuous anisotropic rock, generalized hook s law has been used. Generalized Hooke s law is given by: C.... (2.1) ij ijkl kl where ij and kl denote the stress and strain component, respectively and Cijkl is called the tensor of elastic constants or tensor of compliance. Cijkl has 81 independent components in the most general case but due to the symmetry in stress and strain tensors it can be reduced to 21 components since:,... (2.2) ij ji kl lk C ijkl C... (2.3) klij C ijkl C... (2.4) jikl C ijkl C... (2.5) jilk Equation (2.1) can be rewritten as: A.... (2.6) ij ijkl kl where Aijkl is called the compliance tensor. (Lekhnitskii 1963) used the notation below to show the compliance tensor in its general case:

8 1 yx x, yz x, xz x, xy zx Ex Ey Ez Gyz Gxz Gxy xy 1 zy y, yz y, xz y, xy x Ex Ey Ez Gyz Gxz Gxy x y xz yz 1 z, yz z, xz z, xy y E z x Ey Ez Gyz Gxz G xy z. yz yz, x yz, y yz, z 1 yz, xz yz, xy yz xz Ex Ey Ez Gyz Gxz G xy xz xy xz, x xz, y xz, z xz, yz 1 xz, xy xy Ex Ey Ez Gyz Gxz Gxy xy, x xy, y xy, z xy, yz xy, xz 1 Ex Ey Ez Gyz Gxz G xy... (2.7) where: Ex, Ey, E z are Young s moduli along the directions x,y, and z, respectively Gyz, Gxz, G xy are shear moduli for planes parallel to YZ,XZ,XY planes, respectively yx, zx, zy, xy, xz, yz are Poisson s ration. If the rock is assumed to be compressed (or stretched) along the axial direction, the ratio of the strain in transverse direction to the strain along axial direction is called Poisson s ration. For example, xy indicates Poisson s ration in a plane where there is an axial strain (and axial stress) in X direction and a transverse strain in Y direction. Poisson s ration for this case will be the fraction of strain in Y direction to strain in X direction.

9 ij, kl characterizes the shear stress in the plane parallel to ij-plane that induces k, ij the tangential stress in the plane kl. is called the coefficient of mutual influence of the first kind and shows the stretching/shortening in the direction parallel to the one defined by ij ij, k is called the coefficient of mutual influence of the second kind. It characterizes the shear stress in the ij-plane under the influence of normal stress acting in the k-direction. Note that components of the compliance matrix are usually shown with a ij. For example in (2.7), we have: a12 E yx y The compliance tensor in (2.7) can be simplified using the assumptions below: ij E i ji... (2.8) E j ik, jk jk, ik G jk... (2.9) G ik ij, k k, ij E k... (2.10) G ij The last three equations above simplify the compliance tensor as some of elastic constants vanish and the compliance matrix becomes diagonally symmetrical.

10 2.5 Concept of Transverse Isotropy A special type of orthotropic materials that possess similar mechanical properties in one plane (called plane of isotropy or plane of symmetry) and different properties in normal directions to this plane are called transversely isotropic materials. Considering a coordination system XYZ, the X-axis coincides with the axis of elastic symmetry (Fig. 2). This axis is called axis of elastic symmetry of rotation or axis of rotation. A rock that shows this kind of elastic symmetry is called a transversely isotropic rock. Most of the layered and deposited rocks are characterized as transverse (transversely) isotropic. Fig. 2 shows the axis of rotation (X-axis) and isotropic plane (YZ plane). Fig. 2 Axes orientation in a transversely isotropic material.

11 For rocks with a plane of symmetry as shown in Fig. 2, components of (2.7) reduce to: E E E... (2.11) y z Ex E'... (2.12)....(2.13) yz zy '... (2.14) xz xy E Gyz Gzy G... (2.15) 2(1 ) G G G'... (2.16) xz xy Among the elastic constants above, the shear modulus G can be expressed in terms of Young s modulus E and Poisson s ration,. Therefore, the number of unknowns in a transverse isotropic rock reduces to 5 from total 9 constants in fully anisotropic material. The shear modulus perpendicular to the plane of isotropy, G can also be estimated to reduce the complexity in the problem. Several authors (Batugin and Nirenburg 1972; Cauwelaert 1977; Ellefsen et al. 1992) have worked on experimental methods to estimate the shear modulus G in anisotropic rocks. The relation proposed by (Cauwelaert 1977) is proved to be valid only for rocks with low anisotropy ratio(anisotropy of about 10 percent). (Ellefsen et al. 1992) proposed an experimental procedure for shear modulus estimation in transverse isotropic rocks but their method requires special devices and auxiliary parameters. Among the researches, the relation suggested by (Batugin and Nirenburg 1972) follows relatively simpler analytical

12 solutions and agrees better with experiments. For example, for Y as the axis of rotation, the shear modulus in the XZ-plane is given by: G' G xz Ex. Ez E E 2. E x z xz z... (2.17) To lower the complexity and simplifying the calculations (2.17) is used in this work to estimate G. This reduces the number of elastic properties from 5 to 4. 2.6 Pore Pressure and Effective Stress Pore fluids play an important role in reservoir geomechanics because they support a portion of the total applied stress. Terzaghi s definition of effective stress is: '.... (2.18) ij ij ij p where: ' is the effective stress, ij ij is the total confining stress, ij is the Kronecker delta. p is the pore pressure. Biot suggested an exact effective stress law as below: '..... (2.19) ij ij ij p where, is called Biot effective stress coefficient. It varies between 0 and 1 and describes the efficiency of the fluid pressure in counteracting the total applied stress. Its

13 value depends on the pore geometry and the physical properties of the constituents of the solid system and, hence, on the applied load. For example in sandstones with variable clay content, changes considerably, but due to the results of (Kwon et al. 2001)in shales with extremely high clay content where there is no stiff rock matrix to support externally applied stresses, it seems that applying the simple Terzaghi model (2.18) is more representative. In this study, hydraulic fracturing initiation pressure has been studied regarding the effective stress law mentioned in (2.18). 2.7 Transformation of Elastic Constants As seen in (2.6) and (2.7), to define a relation between stress and strain in a material, we require a compliance tensor. In isotropic materials there is no change in rock s mechanical properties with respect to axes of rotation. But in anisotropic materials, since the elastic constants vary from point to point, rotation of the coordination system has a significant effect on the components of the compliance matrix. To study the effects of different anisotropy orientation and borehole trajectory on compliance tensor, three different coordination systems are required: - X,Y,Z is the global coordination system which in this study is chosen as the orientation of in-situ principal stresses. - X,Y,Z is the coordination system attached to the rectilinear anisotropy. - X, Y, Z is the coordination system attached to the borehole.(local coordination) b b b

14 The reason to define these three different axes is that the orientation of the plane of rectilinear anisotropy and the borehole trajectory do not necessarily coincide the global orientation. So to transform the stresses in global coordination system (principal stresses) and anisotropic elastic constants in borehole coordination system which we are interested to study, compliance tensor transformation calculations are required. As mentioned, to find the stress compliance tensor in the generic form, we need to transform the mechanical properties in anisotropy and borehole coordination systems to a unique system such as the global one. Since in this study, the rock layering is assumed to be horizontal, the rectilinear anisotropy coordinate system coincides with the global one, and simplifies the calculations (Fig. 3). The matrix transformation calculations have been mentioned in Appendix 1. Fig. 3 Position of the horizontal well, local and global coordinate systems attached in a transverse isotropic rock.

15 As a summary, in this section the basic definitions used in linear poro-elasticity are discussed and introduced and the mechanical properties of the rock in anisotropic (transverse isotropic) rocks are investigated. Also the concept of the compliance tensor and constitutive equations are described and the compliance tensor of the rock mass around the borehole in an anisotropic rock with arbitrary (horizontal) orientation was calculated. The calculated compliance tensor enables us to find the stress distribution around the well at every trajectory from principal stresses and with every different anisotropic characteristic.

16 3. ANALYTICAL EQUATIONS FOR FRACTURE INITIATION PRESSURE 3.1 Introduction The objective of this section is to develop an analytical solution for prediction of fracture initiation in a wellbore in transversely isotropic rock (i.e., horizontal drilling through horizontal layers of shales) for different anisotropic ratios. To find the fracture initiation pressure, the stress distribution around the well should be known. The stress distribution around the well in anisotropic rock is highly dependent on the respective angle of the well orientation with the plane of the transverse isotropy, rock mechanical properties and in-situ stresses. The induced components of the stresses around a wellbore are fist calculated and then added to in-situ stresses to find the total stress around the well. The calculated stress distribution is then used in a failure criterion to estimate the hydraulic fracture initiation pressure. To simplify the calculation of the stress distributions around the borehole some assumptions are required: - The rock is a transverse isotropic, in which, the plane of isotropy is horizontal. i.e. a shale reservoir which is horizontally deposited. - The principal stresses are horizontal and vertical and orthogonal to each other. Or in other words, we will have two principal stresses in horizontal plane and the third principal stress in the vertical plane.

17 - The drilled horizontal well trajectory can be along minimum/ maximum principal horizontal stress ( or h H ) or in between them. The procedure to find the hydraulic fracture initiation pressure is as follows: - The stress distribution around a horizontal well in transversely isotropic rock should be calculated. - A proper failure criterion is chosen to assess the fracturing initiation process. - Since in assessing the stress distribution around the wellbore, the stress distribution depends on the angle and also mud pressure inside the well, at any point around the well, the lowest mud pressure that can satisfy the failure criterion is considered as the fracture initiation pressure and the location in which the lowest satisfying mud pressure has been obtained is considered as the fracture initiation angle. The general expressions for calculation of the stress distribution around the wellbore are discussed below. 3.2 General Expressions for Calculation of the Stress Functions To find the stress distributions at any point in an elastic anisotropic rock, five different sets of equations must be satisfied: 1. Equation of Equilibrium 2. Strain- Displacement Relations

18 3. Compatibility Equations for Strains 4. Constitutive Equations 5. Rocks Boundary Conditions To find the stress state at any point in an elastic anisotropic rock, we need to use equations 1 to 4 and check the results in the 5 th equation (boundary conditions). These sets of differential equations are called Beltrami-Michell equations of compatibility for anisotropic rocks it can be written as below: L F... (3.1) 4 L 3 0 L F... (3.2) 3 L 2 0 where: F and are the stress functions at a point, And L2, L3, L4are the linear differential operators and can be calculated as below: L 2 2 2 2 x xy y 2 44 2 45 55 2... (3.3) 3 3 3 3 L ( ) ( ) x x y xy y 3 24 3 25 46 2 14 56 2 15 3......(3.4) L 4 4 4 4 4 2 2 (2 ) x x y xy x y y 4 22 4 26 3 16 3 12 66 2 2 11 4... (3.5) where: a. a i3 j3 ij aij ( i, j 1,2,4,5,6)... (3.6) a33

19 And aij are components of the compliance tensor, defined in previous section. If we solve the equations above in terms of F, the equation below is achieved: ( L L L ) F 0... (3.7) 2 4 2 3 According to (Amadei 1983), L3 0 corresponds to the case where there is a plane of elastic symmetry perpendicular to the borehole axis. According to (Amadei 1983), the algebraic equation that corresponds to equation (3.7) can be written as: l ( ). l ( ) l ( ) 0... (3.8) 2 4 2 3 where: l ( ) 2... (3.9) 2 2 55 45 44 l ( ) ( ) ( )... (3.10) 3 2 3 15 14 56 25 46 24 l ( ) 2 (2 ) 2... (3.11) 4 3 2 4 11 16 12 66 26 22 Equation (3.8) always has 6 roots of ( i 1 6). (Lekhnitskii 1963) analytically showed that the roots are always purely imaginary and complex and also showed that three of the results are always conjugate roots of three others. So he ignored the conjugate roots and took the first third roots as the answer of the equations. So, to be able to calculate stress distribution in the cases of L3 0we need to calculate ( i 1 3). (Amadei 1983) mentioned 4 different conditions of anisotropy for which the plane of symmetry is perpendicular to the borehole axis and called them special cases of anisotropy : i i

20 1- Orthotropic material with one elastic symmetry plane perpendicular to the well axis and two other planes parallel to the well axis 2- Transverse isotropy in a plane perpendicular to the well axis 3- Transverse isotropy in a plane striking parallel to the well axis 4- complete isotropy Fig. 4 and Fig. 5 show the cases 2 and 3 of the four stress conditions Amadei has mentioned for analytical solutions mentioned above. Fig. 4 Transverse isotropy in a plane perpendicular to the well axis.

21 Fig. 5 Transverse isotropy in a plane parallel to the well axis. Among these 4 cases, case 2 is the one that can model horizontal drilling in a horizontally deposited shale formation. (Ong 1994) used the methods proposed by (Amadei 1983) and (Lekhnitskii 1981) and suggested a series of equations for stress distribution around the wellbore and in this study his contributions will be used. 3.3 Stress State around the Wellbore Generally, the final state of the stress around a wellbore is sum of the in-situ stress (original, xo, ) and the stress induced by drilling or excavation (, xi ). Each of these will affect the total stress distribution; however, we will focus on the induced

22 stresses in this work and will not discuss the impact of rock anisotropy on the in-situ stress state. 3.3.1 Wellbore Stresses in Transversely Isotropic Rock in Generic Form The studies to find the stress distribution around a wellbore in an anisotropic rock are done using either analytical or numerical methods. For example, (Suarez-Rivera et al. 2006) used finite element modeling to estimate the stresses around a borehole drilled in transversely isotropic rock and extracted approximate expressions for the tangential stress around the hole. As mentioned before, the analytical approaches are based on the theoretical works of (Lekhnitskii 1981) and (Amadei 1983). Similarly to (Ong and Roegiers 1993) and (Ong 1994), in this study, the analytical work of (Amadei 1983) was used to derive expressions for induced stress distribution around the well and for different borehole orientations of interest. The total stress components around the wellbore wall are given by (Amadei 1983) as the sum of the original (in-situ) stress (shown by o subscript) and a term for an induced component of the stress (shown by i subscript) as follows:... (3.12) x x, o x, i... (3.13) y y, o y, i... (3.14) xy xy, o xy, i

23... (3.15) xz xz, o xz, i... (3.16) yz yz, o yz, i... (3.17) z z, o z, i The first term in these equations is the original (in-situ) stress (which should be measured or estimated separately and is assumed known) and the second term is the induced component of the stress. For induced component of the stresses, (Amadei 1983) proposed: 2 2 2 x, i 2e 1 1 z1 22 z2 3 33 z 3... (3.18) yi, 2e 1 z1 2 z 2 33 z3... (3.19) xy, i 2e 1 1 z1 22 z2 3 33 z 3....(3.20) xz, i 2e 1 1 1 z1 222 z2 33 z 3... (3.21) yz, i 2e 1 1 z1 22 z 2 3 z3....(3.22) 1 a a a a a... (3.23) z, i 31 x, i 32 y, i 34 y, i 35 xz, i 36 xy, i a33 In the above expressions: - e : is real component of an complex expression - 1 is the positive root of the equation: 2 0... (3.24) 2 55 45 44, are the positive roots of: - 2 3

24 2 (2 ) 2 0... (3.25) 4 3 2 11 16 12 66 26 22 Equations (3.24) and (3.25) are derived from the sixth order equation, shown in (3.4). The results of i are pure imaginary numbers and half of the roots are conjugates of the other half. - ij is called the reduced strain coefficient and can be calculated using (3.6) - aij are components of the compliance tensor, defined in the previous section. -, i 1,2,3 are complex numbers defined by (Amadei 1983)and can be i calculated using: l ( ) 3 i i i 1,2,3... (3.26) l2( i ) where the functions l 3 ( i ) and l 2 ( ) are introduced in (3.10) and (3.9) respectively i - (Amadei 1983) proposed that the terms ', i 1, 2,3 are derivatives of the stress functions and are calculated using equations below: i 1 ' ( z ) [( i P )( ) 1 1 xy, o yy, o w 2 2 3 3 z1 2 2 2* * 1* ( ) 1 1 a ( i ip )( 1) ( i ) ( )] xy, o xx, o w 2 3 yz, o xz, o 3 3 2... (3.27) 1 ' ( z ) [( i P )( ) 2 2 xy, o yy, o w 1 3 3 1 z2 2 2 2* * 2* ( ) 1 2 a ( i ip )(1 ) ( i ) ( )] xy, o xx, o w 1 3 yz, o xz, o 3 1 3... (3.28)

25 1 ' ( z ) [( i P )( ) 3 3 xy, o yy, o w 2 1 1 2 z3 2 2 2* * 3* ( ) 1 3 a ( i ip )( ) ( i )( )] xy, o xx, o w 1 2 yz, o xz, o 2 1... (3.29) - The complex number Z i in the above equations is defined as: Z., 1,2,3 i x i y i... (3.30) where x and y are the coordinates of the points of interest around the wellbore. In polar coordinate system Zi can be written as: Z a(cos.sin )... (3.31) i i where a is the well radius and is the angle measured counterclockwise from the horizontal (x) direction (show in Fig. 6). Fig. 6 Transformation around the wellbore from rectangular to polar system.

26 - For parameters and i we have: ( ) ( )... (3.32) 2 1 2 3 1 3 1 3 3 2 i zi a zi ( ) 1 a 1 i 2 2 i i, i 1, 2,3... (3.33) Using (3.31) and at the wellbore wall (r=a), it can be mathematically shown that: i i e... (3.34) If we substitute (3.31) and (3.34) in equations of derivatives of the stress functions eqns. 25-27, we simplified the derivatives of the analytic function ' ( i ) as below: ( z ) A[ B( 1) C( ) D( ) ]... (3.35) ' 1 1 i 2 3 2 2 3 3 3 2 3 ( z ) A [ B(1 ) C( ) D( ) ]... (3.36) ' 2 2 2 1 3 1 3 3 1 1 3 3 ( z ) A [ B( 1) C( ) D( ) ]... (3.37) ' 3 3 3 2 3 2 2 3 3 3 2 3 where: A i 1... (3.38) 2 ( cos sin ) i B ( p )cos sin i[( p )sin cos ]... (3.39) w x, o xy, o w x, o xy, o C ( p )sin cos i[( p )cos sin ]... (3.40) w y, o xy, o w y, o xy, o D cos sin i[ cos sin ]... (3.41) xz, o yz, o yz, o xz, o

27 So in the generic form, to find the stress distribution around a well in an anisotropic (transversely isotropic) rock, a process should be followed and the parameters below should be calculated: (a) The in-situ stresses are specified. If the wellbore orientation is different from the orientation of the principal stresses, shear stresses also will exist and stress transformation relations (explained in previous section) should be used. (b) Mechanical properties of the rock (including Young s moduli, Poisson s ration and shear moduli) should be specified in XYZ directions and the compliance tensor should be formed. (according to equation (2.7) in SECTION 2) (c) Using components of the compliance tensor (a ij ) in (2.7) and (3.6), the parameter ij is calculated. (d) Using (3.24) and (3.25), the values of ij (that is already calculated), the values,, are calculated. of 1 2 3 (e) The, i 1, 2,3 are used in (3.26) to find, i 1,2,3 i i (f) The value of is calculated using (3.32) (g) The derivatives of the analytic function ' ( i ) are calculated using eqns. 33-39. (h) The parameters obtained in (a-g) are finally plugged into eqns. 14-19 to give the induced components of the stresses.

28 (i) Induced stresses are added to original (in-situ) components of the stresses to give the total stress distribution around the well. 3.3.2 Wellbore Stresses in Transversely Isotropic Rock Drilled along Principal in-situ Stresses In SECTION 3.3.1, the general solutions for assessment of stress distribution around a horizontal well in horizontally bedded rocks are explained. Since drilling along the in-situ principal stresses is quite common in petroleum industry i.e., for achieving the best fracture shape (longitudinal or transverse fractures), borehole stability issues and minimizing drilling related problems, in this section, the equations mentioned above will be reduced to find the stress distribution around a horizontal well drilled along one principal in-situ stress. The results are then used to estimate the hydraulic fracture initiation pressure around the wellbore in horizontal wells drilled along a principal stress. By assuming that the horizontal well is drilled along one of the principal horizontal stresses, or, the far-filed shear stress components vanish: h H... (3.42) xy, o yz, o xz, o 0 Also, for horizontal wellbores drilled in horizontally laminated transversely isotropic reservoirs, consider that the plane of isotropy is the horizontal plane XZ and the axis of symmetry is the Y-axis. Also, let the principal stresses be vertical and horizontal and aligned with the global coordinate system XYZ and the wellbore is attached to the

29 local coordination system as shown in Fig. 7 (Y b is vertical, while X b and Z b are horizontal). Fig. 7 Horizontal well orientation with respect to 3 coordination systems. By solving (3.24) and (3.25) for, i 1,2,3 and substitution in equation 22-24, we get: i 0, i 1, 2,3... (3.43) i Using (3.43) in (3.32) we get:... (3.44) 2 1

30 Also using eqns. 40-42 in eqns. 33-39, we simplified the values for derivatives of the stress analytical functions ' i as follows: ' ( z ) A.[ B C ]... (3.45) 1 1 1 2 ' ( z ) A.[ B C ]... (3.46) 2 2 2 1 ' 3 0... (3.47) where A i is defined in (3.38) and B, C have been simplified to: B ( p )cos i[( p )sin ]... (3.48) w x, o w x, o C ( p )sin i[( p )cos ]... (3.49) w y, o w y, o By applying the solutions and simplification mentioned in eqns. 42-47, the stress distribution on a wellbore along the principal stresses can be calculated as illustrated in eqns. 48-53. Assuming that the wells have been drilled along maximum horizontal stress, the X, Y and Z directions should be along the, and v h H stresses, respectively. For total stresses we will have: 2 2 x x, o x, i v 2e 1 1 z1 22 z 2... (3.50) y y, o y, i h 2e 1 z 1 2 z2... (3.51) 1 a a... (3.52) z H 31 x, i 32 y, i a33 xy xy, i 2e 1 1 z1 22 z 2... (3.53)

31 0... (3.54) xz 0... (3.55) yz In these equations, the subscripts i and o stand for induced and original components of the stresses, respectively. The stress values are in Cartesian coordination should be converted to polar coordination to be used in borehole geometry. It will be assessed more in SECTION 4. 3.4 Assessment of a Proper Failure Criterion for Fracturing Analysis Different failure criteria are used to assess the hydraulic fracturing initiation. These criteria are for shear and tensile failure criteria. Shear failure is more common in wellbore stability analysis. However, tensile strength seems to be the best type of criterion for assessing the hydraulic fracture initiation in the wells since rocks have much lower tensile strength than shear strength, so they can break in tension easily. (Daneshy 1973) in his paper presented a criterion for tensile fracturing in an inclined well where the borehole axis has an angle with the orientation of the principal stresses. He considered the effects of the fluid invasion inside the permeable zone on the variation of the induced stresses. But his solution was for isotropic and so not applicable to anisotropy. (Hossain et al. 2000) also studied the hydraulic fracture initiation pressure in inclined wells but he also considered the rock as isotropic.

32 In this study, after finding the stress distribution around a wellbore in transversely isotropic rock, tensile strength criterion was used to assess the breakdown pressure in the wells. Tensile failure criterion of the rock states that the fracture initiates where and when the effective tangential stress magnitude equals to rocks tensile strength. In other words:... (3.56) P p t For simplicity, one can assume that the pore pressure at the wall equals the mud pressure ( p w). By calculating the wellbore pressure ( pw) that satisfies the tensile strength criterion, the location and position of fracture initiation around the well can be estimated. Fracture initiation pressure is the pressure at which p w overcomes the tangential stress at the wellbore s weakest point. Since the hoop (tangential) stress is dependent on the angle, to find the weakest point around the well, (3.56) should be checked for all angles between 0 to 360 degrees, each yielding a different value of p w pressure. Then, the minimum value of pw is the fracturing pressure (breakdown pressure) at the weakest point around the well which is the fracture initiation point (point of lowest hoop stress). As mentioned above, to find the breakdown pressure ( p wb ), first the tangential stress or which is function of the mud pressure inside the well ( p w ) is calculated. Since the pore pressure and the rock tensile strength are already known, the critical mud pressure that can satisfy the equation above (the breakdown pressure) can be calculated.

33 So, as the summary, in this section the general procedure for calculation of stress distribution around a horizontal well drilled inside a horizontally bedded transversely isotropic rock was explained. Also the equations are simplified for the case of a horizontal well drilled along maximum horizontal stress ( ). At the end, to validate Hmax the proposed equations, the anisotropic solution was used to find the stress distribution for a nearly isotropic rock and the results are compared with classic isotropic solutions proposed by Kirsch and a perfect match was reported. In next section, the stress distribution equations will be used to find the breakdown pressure in rocks with different anisotropy ratios. 3.5 Estimation of Fracture Initiation Pressure Fracture initiation pressure is an important parameter in petroleum industry since it is used in pre-drilling analysis, wellbore stability, and safe mud weight design and should be known to avoid drilling problems such as ballooning, lost circulation and so on. As mentioned in the literature review, many different equations have been proposed to help estimate the formation fracturing pressure before drilling. One of the most common equations is the equation proposed by (Hubbert and Willis 1957). They used the linear isotropic assumption and assumed the well as vertical. In this study our results will be compared with the results of (Hubbert and Willis 1957). Their proposed equation for fracture initiation pressure in a vertical well is as follows: P 3... (3.57) b h H p t

34 Since in horizontal wells, the maximum and minimum in-plane stresses are different from vertical wells, equation of (Hubbert and Willis 1957) is applied to a horizontal well drilled along H max as follows: P 3... (3.58) b h v p t Equation (3.58) will be used in next sections to verify the results of fracture initiation pressure with anisotropic solutions.

35 4. CALCULATION OF FRACTURE INITIATION PRESSURE IN A HORIZONTAL WELL DRILLED IN SHALE 4.1 Introduction In SECTIONS 1 to 3, the basics of anisotropy, analytical solutions for calculation of stress distribution around the well in different orientations and proper failure criterion for fracturing purposes are explained. In this section, using the equations mentioned in previous sections, the isotropic solutions are compared with anisotropic ones to validate the proposed equations. Numerical examples are solved to show the applications of the proposed equations. Sensitivity analysis is also carried out to assess the effects of anisotropy (Young s modulus and Poisson s ration) on fracture initiation pressure. 4.2 Numerical Example to Find Tangential Stress around the Well In this section a numerical example is shown to illustrate the procedure for calculation of the tangential stress in horizontal wells. Since the case of drilling along the principal stresses comparatively requires fewer calculations, it was chosen as the numerical example. Consider a horizontal well in shale under an in-situ stress and mechanical properties given in Table 1 and Table 2 for a depth of 10,000 ft. (3048 m) and Table 1 shows available in-situ stresses and rock mechanical properties for calculations.

36 Table 1 Available in-situ Stresses and Mechanical Properties for Calculations σ y =σ v σ x =σ h σ z =σ H σ p σ t psi/ft psi/ft psi/ft psi/ft psi pa/m pa/m pa/m pa/m Mpa 1 0.8 0.85 0.61 100 (22.6x10 3 ) (18.09x10 3 ) (19.23x10 3 ) (13.79 x10 3 ) (0.69) ft. (3048 m). Fig. 8 shows the coordination systems and in-situ stresses at the depth of 10,000 Fig. 8 Well position with respect to in-situ stresses and the coordination system.

37 In this example, the mechanical properties used in the calculations are chosen close to each other to make the rock nearly isotropic. The fracture initiation angle and pressure are calculated using transversely isotropic solutions. They are then compared with isotropic solutions. This step can be considered as the verification of the transversely isotropics solution as well. It should be noted that nearly isotropic material properties are used and not the exactly isotropic because in the later case, the anisotropic solution fails. Because using exactly similar rock mechanical properties results 1 i and 2 3 i in (3.24) and (3.25). Now if the results of ', i 1,2 are calculated using i equations (3.45) and (3.46) and used in equations (3.50) and (3.51), it can be seen that the right hand side expression which denotes the induced component of the stresses, is zero. Therefore, using exact rock mechanical properties causes wrong total stress values along X and Y directions. Table 2 Mechanical Properties of the Rock Used to Validate the Solutions E hor =E x =E z (psi) E vert =E y (psi) Gyz=Gxy (psi) Gxz (psi) ν hor =ν xz ν vert =ν yz =ν xy 1.40001x10 6 1.4x10 6 5.60002x10 5 5.60008x10 5 0.24999 0.25 The compliance tensor of the rock with properties in Table 2 is:

38 1 yx zx 0 0 0 Ex Ey Ez xy 1 zy 0 0 0 Ex Ey Ez xz yz 1 0 0 0 Ex Ey E...(4.1) z 1 0 0 0 0 0 G yz 1 0 0 0 0 0 Gxz 1 0 0 0 0 0 Gxy 7 7 7 7.142806*10 1.78571*10 1.78563*10 0 0 0 7 7 7 1.785701*10 7.142857*10 1.785701*10 0 0 0 7 7 7 1.785630*10 1.785714*10 7.142806*10 0 0 0 6 0 0 0 1.785709*10 0 0 6 0 0 0 0 1.78568*10 0 0 0 0 0 0 1.785709*10 6 The matrix Beta can then be calculated using the compliance tensor in (4.1) and (3.6): ij 7 7 6.69641*10 2.23212*10 0 0 0 0 7 7 2.232109*10 6.69643*10 undefined 0 0 0 undefined undefined undefined 0 0 0 0 0 0 6 1.785709*10 0 0 0 0 0 0 6 1.785687*10 0 0 0 0 0 0 1.785709*10 6... (4.2)

39 In equation above, the components a and i3 a, 1,2,3 3 i i are undefined since equation (3.6) is not defined for i, j 3 because it causes a zero value for ij in equation (3.6) which makes equation 3.9 to 3.11 unsolvable. Using (3.24) and (3.25) (in SECTION 3) and the values of ij (already calculated in (4.2)), the values of 1, 2, 3 are calculated. (,, ) (0.999995 i,1.0000057 i,1.0000061 i)... (4.3) 1 2 3 According to SECTION 3, 0, i 1,2,3... i Also 2 1 0.000010391i... (4.4) (4.5) ' ( z ), i1,2,3 is calculated using equations 36 and 43-47 in SECTION 3. Before i i calculation of ' ( z ) i i, we need to calculate the parameters A i, B, C as below: A i 1 6 20.782*10 i*( i cos sin )... (4.6) B p i p... 3 3 ( w 8*10 )cos [( w 8*10 )sin ] (4.7) C p i p... (4.8) 4 4 ( w 10 )sin [( w 10 )cos ] Using eqns. 43-45 (in SECTION 3) and after calculating variables A i, B, C for calculation of ' ( z ) and after some mathematical simplifications we have: i i ' 2i 7 1( z1) e (9.624 10 0.27750 p w )... (4.9) ' 2i 7 2( z2) e ( 9.624 10 0.22250 p w )... (4.10)

40 ( z ) =0... (4.11) ' 3 3 So by plugging the above mentioned variables in eqns. 48-49 (in SECTION 3) we will get: p 2 3 2i w x 810 2 ee [ ]... (4.12) y p 2 4 2i w 10 2 ee [ (5000 )]... (4.13) And using (3.53) in SECTION 3 for xy calculations we have: xy p e[sin 2 ]... (4.14) w Now that all the required variables are calculated, the hoop stress ( ) for the case of a nearly isotropic rock can be found using equation below:... (4.15) 2 2 sin. x cos. y sin 2. xy 2pp t Equation above is calculated using a simple conversion from Cartesian to polar coordination system. As seen above, the equations are functions of the angle around the well ( ) and the mud pressure inside the well ( p w). In the next section, as an example, the hydraulic fracture initiation pressure is calculated using anisotropic solutions and the results are compared with the classic isotropic solutions.

41 4.3 Comparison around the Well Using Isotropic and Anisotropic Solutions (Kirsch 1898) proposed a set of equations to calculate the stress distribution around a vertical well. Using (Kirsch 1898) equations and by changing the vertical well assumption to horizontal well (by changing the term H to v as the maximum in-plane stress in normal fault regime zones) and some simple mathematical manipulations, the tensile stress distribution around a horizontal well can be calculated as follows: 2 4 2 1 a 1 3 a p * a ( v h 2 Pp )(1 ) ( )(1 )cos 2 2 v h 4 2 t... (4.16) 2 r 2 r r At wellbore wall r=a, and it simplifies to: 2( )cos2 2P... (4.17) h v v h p t Since due to the maximum tensile strength criterion, the fracturing initiates at the point with the minimum hoop (tangential) stress values, therefore, the fracture initiates in a horizontal well at the points with minimum tangential stress as below: 3 2P p... (4.18) min h v p t where p shows the pressure difference between the mud weight and pore pressure. Using the input examples mentioned in Table 1 and Table 2 in Fig. 9 the tangential stress is calculated with isotropic and anisotropic solutions and is compared to each other. These calculations are done with assuming the mud pressure of 8000 psi.

42 Fig. 9 distribution around a nearly isotropic horizontal well drilled along H max and with mud pressure of 8000 psi using isotropic and anisotropic solutions. As can be seen in figure above, at 90,270 which coincides the orientation of the maximum in-plane stress (upper most and lower most points of the well), the tensile strength has its minimum value and is the weakest point around the well for tensile fracturing. So at 8000 psi mud pressure inside the well, the tangential stress goes to tension at 90,270 and the fracture initiates at these two points. Fig. 9 shows how well the results of isotropic and anisotropic solutions match with each other for a nearly isotropic rock and can be considered as the validation of the solutions.

43 4.4 Comparison around the Well Using Different Anisotropy Ratios In this section, to show how rock anisotropy can change the stress distribution around the well, three cases of Young s modulus anisotropy are selected and the tangential stress distributions around the well in these three cases are compared. In these cases only the effects of Young s modulus have been studied and the Poisson s ration was assumed isotropic since it is assumed that Poisson s ration doesn t contribute significantly on changing the hoop stress around the well. These cases are: Ehor 1. A nearly isotropic case, where 1 E Evert 2. Anisotropic case where 1.6 E hor Ehor 3. Anisotropic case where 1.6 E vert vert Fig. 10 shows the tangential stress distribution around the well for cases 1 to 3 mentioned above. In these cases, the mud pressure inside the well is assumed to be 8000 psi.

44 Fig. 10 Tangential stress distribution around the well for three cases with different anisotropy ratios. As can be seen in figure above, at 8000 psi mud pressure, the isotropic rock ( E E hor vert 1or case 1) is about to fail at 90 and 270 degrees since at these two points, the tangential stress curves have zero values. For case 2 where Evert 1.6, the tangential stress E increases and the tensile fracture can t initiate in this case. In other words, to initiate the Evert tensile fracture in case of 1.6, we require the mud pressure slightly more than E 8000 psi. hor hor

45 Ehor For case 3 where 1.6, as can be seen in Fig. 10, the tangential stress has E vert gone to tension since its values has become negative in a big zone. Therefore, at 8000 psi mud pressure, the fracture has already been initiated in this case. So, it is obviously seen that Young s modulus anisotropy can increase or decrease the minimum hoop stress- the parameter that plays a key role in hydraulic fracture initiation. 4.5 Estimation of Fracture Initiation Pressure in Horizontal Wells Drilled in Horizontally Deposited Rock In this section, the effect of different rock properties (different Young s modulus and Poisson s ration) on changing the value of the fracturing initiation pressure is discussed. The in-situ stress conditions are similar to the case introduced in Table 1 in SECTION 3. Rock mechanical properties used for calculations are also mentioned below in Table 3. It is assumed that the well is drilled along maximum horizontal stress ( H max ) direction. A sensitivity analysis was carried out to show how variations in rock mechanical properties change the values of fracture initiation pressure. As can be seen in Table 3, the Young s modulus and Poisson s ration in vertical 6 direction was assumed constant ( E 1.4 10 and 0.2 ) and then Young s vert modulus in horizontal direction changes between 0.5 E E 3 E and the Poisson s ration in horizontal direction between 0.05 hor 0.45. vert vert hor vert

46 Due to equation E G 2(1 ) in the isotropic (horizontal) plane, assuming Poisson s ration as constant, there is a linear relationship between Young and shear moduli and by changing 0.5 E E 3 E, the shear moduli will also change to 0.5G G 3G vert hor vert vert hor vert Table 3 Mechanical Properties of the Rock Being Used for Sensitivity Analysis ν E vert =E y E hor =E x =E z G yz =G xy G xz ν hor =ν vert =ν yz =ν x xz (psi) (psi) (psi) (psi) (fraction) (fraction) y 1.4x10 6 (0.5-3)x1.4x10 6 5.6x10 5 (0.5-3)x5.6x10 5 (0.05-0.45) 0.2 Table 4 shows some numerical examples of how Young s modulus and Poisson s ration can change the value of fracturing pressure in a horizontal well drilled along the maximum horizontal stress direction. 6 numerical examples are shown in Table 4 with different Young s modulus and Poisson s ration values. As can be seen, the Ehor hor value of fracturing pressure when 1equals to 8000 psi which matches with E results of isotropic calculations (equation (3.58)) where: vert vert P 3 38000 10000 6100 100 8000 psi... (4.19) b h v p t

47 Depending on the Young s modulus and Poisson s rations, the fracture initiation pressure can be more or less than the fracturing pressure in isotropic case. For example, Ehor hor when Young s modulus is 0.5and Poisson s ration is 1, the fracturing E vert pressure changes from 8000 psi in isotropic to 8877 psi in transversely isotropic rock. vert Table 4 Numerical Example to Show Variations of Fracturing Pressure with Different Young s Modulus and Poisson s Ration Young s Modulus E Poisson Ratio hor vert 0.5 1 1.5 0.5 8845 psi 8030 psi 7650 psi 1 8877 psi 8000 psi 7726 psi 1.5 8914 psi 8153 psi 7835 psi E hor vert To further study the effects of anisotropy on fracturing pressure, 3000 data points are created using different combinations of Young s modulus ratio E E hor vert and Poisson s ration hor vert. These points are selected using the criteria defined in Table 1. The selected data points are then used in the proposed analytical solutions mentioned in SECTION 3.3.1 and 3.3.2 to find the stress distribution around the well and finally the required mud pressure to initiate the first tensile fracture (fracture initiation pressure) was calculated separately for each case. The calculated fracture

48 initiation pressures are plotted in Fig. 11 and Fig. 12 for the wells drilled along H max and hmin respectively. These plots show the variations in fracturing pressure caused by changes in the mechanical properties. Fig. 11 Fracturing pressure variations in a well drilled along Young s modulus and Poisson s rations. H max by changing

49 Fig. 12 Fracturing pressure variations in a well drilled along Young s modulus and Poisson s rations. hmin by changing There are several points in Fig. 11 and Fig. 12 which should be noted: The results of the sensitivity analysis clearly show dependency of fracture initiation pressure to variations in E E hor vert and hor vert. Depending on the values of Young s modulus and Poisson s ration, the values of fracture initiation pressure may change. The case of isotropy is where there is no difference between mechanical properties in different directions. So, in Fig. 11 and Fig. 12, the points of

50 E E hor vert hor vert 1 show the pressure required to initiate the tensile fracture (fracture initiation pressure) in the isotropic rock. In Fig. 11 and Fig. 12, the contour lines are highly inclined showing that Poisson s ration doesn t significantly contribute in changing the fracture initiation pressure, but at higher Young s modulus ratios, Poisson s ration is also effective in changing the fracturing pressure. Unlike the Poisson s ration, the Young s modulus ratio changes the fracturing pressure effectively and significantly. For example, for Fig. 11, it can be hor obviously seen in Table 4 that when the Poisson s ration is constant at 1, variation of Young s modulus ratio from 0.5 to 1.5 causes about 1151 psi (8877-7726=1151 psi) difference in the value of fracture initiation pressure. However, if Ehor the Young s modulus is kept constant at 1, variations of Poisson s ration E from 0.5 to 1.5 can only cause 123 psi (8153-8030=123 psi) difference in the fracturing pressures. So, the Young s modulus ratio obviously has higher importance in changing the fracture initiation pressure than the Poisson s ration. vert vert As can be seen in Fig. 11 and Fig. 12, the contours of fracturing pressure variations are much denser in the case of drilling along H max than drilling along

51 hmin and it indicates that rock anisotropy has higher effect on changing the fracturing pressure values during drilling along H max than along hmin. Due to fundamentals of rock mechanics, the materials stiffness has a direct relation with the Young s modulus and an inverse relation with the Poisson s ration. In other words, materials with higher Young s modulus and lower Poisson s ration are respectively stiffer materials (i.e. granite) and those with lower Young s modulus and higher Poisson s ration are softer rocks (i.e. shale).therefore, since in Fig. 11 and Fig. 12, the lower right corner of the plot has high E E hor vert and low hor vert, it represents the case where the rock mass is stiffer in horizontal direction. However, the upper left corner in Fig. 11 and Fig. 12 corresponds to the case where the vertical direction has higher stiffness. The results of sensitivity analysis in Fig. 11 and Fig. 12 indicate that in normal stress faulting regime where v H max hmin and drilling along maximum and minimum horizontal stresses, higher stiffness in vertical direction causes more resistance against fracturing and delays the hydraulic fracturing initiation from the well but higher rock mass stiffness in horizontal direction (having respectively lower stiffness in vertical direction) can cause fracturing earlier than the case of isotropic rock. The analysis gives a good tool to predict in-situ rock mechanical properties by knowing the fracturing pressure in the field and doing the back analysis. For

52 hor example in Fig. 11 and Fig. 12, by assuming 1(for simplicity) and by vert finding the fracturing pressure from the plot, in-situ E E hor vert can be estimated. To better illustrate the importance of the findings, in this study, the fracturing pressure values obtained from those 3000 points in Fig. 11 are converted to equivalent mud weight at depth of 10,000 ft. (because the well is assumed to be drilled at 10,000 ft. depth). For example, knowing that the fracturing pressure in Fig. 11 for isotropic rock is 8000 psi, the mud weight that can cause hydraulic fracture initiation in the isotropic case can be calculated using the equation below: Pw(psi) 8,000 psi Mw(ppg)= = =15.4 ppg... (4.20) 0.052D(ft) 0.05210,000 So, it can be said that using 15.38 ppg mud weight in the assumed well can initiate the fracturing around the well. Fig. 13 shows the equivalent mud weight that can cause fracturing in rock mass when the well is drilled along H max with different mechanical properties.

53 Fig. 13 Variations in equivalent mud weight required to cause fracturing in rocks with different anisotropy ratios. In Fig. 13, it should be noted that: Ehor hor The isotropic case with 1shows equivalent mud weight of 15.4 ppg E vert vert As the rock mass stiffness decreases in vertical direction (respective stiffness increases in horizontal direction), which corresponds to lower right corner of Fig. 13, the mud weight required to initiate the fracturing decreases. In other words the rock will break down earlier than expected. For example, if the required

54 mud weight to cause fracturing in isotropic rock is ~15.4 ppg, for a rock with E E hor vert hor 1.5 and 1, the mud weight falls to 14.85 ppg. vert If rock mass stiffness increases in vertical direction (respective stiffness decreases in horizontal direction), corresponding to upper left corner of Fig. 13, the mud weight required to initiate the fracturing increases. In other words the rock may withstand more pressure without fracturing. For example, for a Ehor rock with 0.5(twice higher Young s modulus in vertical direction than E vert hor horizontal direction) and 1, the fracturing mud weight raises to ~17 ppg. vert So, as an important result, depending on the respective higher stiffness direction in the transversely isotropic rock, the fracturing pressure and equivalent mud weight can be higher or lower than the fracturing pressure expected with isotropic solutions. But since in bedded rocks, the rock stiffness is usually higher in horizontal direction, it is expected that the actual fracturing pressure (calculated from anisotropic solutions) is less than the expected value calculated by isotropic solutions. As another study to show the effects of rock mechanical properties on changing the hydraulic fracture initiation pressure, equation (2.15) and (2.17) were used to calculate shear modulus anisotropy ratio G G hor ver for the horizontal well drilled along and using the Young s modulus and Poisson s ration data in Table 3. H max

55 In this case also, 3000 data points were produced and the fracturing pressure values were calculated for each case and the results are plotted in Fig. 14. Fig. 14 Fracturing pressure variation by changing shear modulus and Poisson s ration. As can be seen in the figure above, the same trend between the Young s modulus anisotropy ratio exists for the shear modulus ratio but the effects are less intense. Similar to Young s modulus anisotropy vs. Poisson s ration anisotropy results shown in Fig. 11, Poisson s ration has a weak effect on changing the fracture initiation pressure while shear modulus changes the fracturing pressure significantly.

56 The other important difference is that since the Poisson s ration changes between 0 and 0.5 and due to (2.15), in the isotropic plane, the shear modulus should change between 1/3 to 2/3 of the Young s modulus ratio. So, it can explain the reason why there are no data points for G G hor ver less than 0.5 and after 2.5. 4.6 Direct and Indirect Estimation of Rock Anisotropy using Experimental Tests As seen in previous section, estimation of hydraulic fracture initiation pressure is related to an estimate of Young s modulus ratio E E hor vert and Poisson s ration hor vert and the expected fracturing pressure (or equivalent fracturing mud weight) changes by variation in rock anisotropy ratio in shales. To be able to find the Young s modulus and Poisson s ration anisotropy values we need to use experimental tests to measure Young s modulus and Poisson s ration in 2 orthogonal directions. In this study, to be able to estimate rock anisotropy, Pierre shale rock was prepared and two sets of core samples were drilled in two different directions: - Parallel direction to the rock bedding - Perpendicular direction to the rock bedding Pierre shale is a geologic formation from upper Cretaceous and expands from North Dakota to New Mexico. It is described as dark gray shale and fossiliferous. The Pierre shale is known for its extremely low permeability but natural fractures improve its

57 in-situ permeability in some areas and make it commercially attractive for gas production. (Wikipedia 2010) Fig. 15 shows the coring and rock bedding directions graphically. Fig. 15 Respective orientation of rock bedding and the core samples in Pierre shale rock. The core samples were tested using triaxial testing machine to measure the Young s modulus and Poisson s ration in orthogonal directions directly. Fig. 16 shows the comparison of Young s modulus for the core sample parallel to the rock bedding with the one being cut in perpendicular direction to the bedding.